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 October 6th, 2012, 11:01 AM #1 Newbie   Joined: Oct 2012 Posts: 13 Thanks: 0 Simplify trig equation cot(2arcsinx) Please help, thanks in advance
 October 6th, 2012, 11:07 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,206 Thanks: 513 Math Focus: Calculus/ODEs Re: Simplify trig equation Using the double-angle identity for cotangent, we may write: $\cot$$2\sin^{\small{-1}}(x)$$=\frac{\cot^2$$\sin^{\small{-1}}(x)$$-1}{2\cot$$\sin^{\small{-1}}(x)$$}$ $\cot$$\sin^{\small{-1}}(x)$$=\frac{\sqrt{1-x^2}}{x}$ Can you finish?
October 6th, 2012, 11:14 AM   #3
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Re: Simplify trig equation

Quote:
 Originally Posted by MarkFL Using the double-angle identity for cotangent, we may write: $\cot$$2\sin^{\small{-1}}(x)$$=\frac{\cot^2$$\sin^{\small{-1}}(x)$$-1}{2\cot$$\sin^{\small{-1}}(x)$$}$ $\cot$$\sin^{\small{-1}}(x)$$=\frac{\sqrt{1-x^2}}{x}$ Can you finish?
Would you just multiply cot(arcsinx) by 2?

 October 6th, 2012, 11:18 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,206 Thanks: 513 Math Focus: Calculus/ODEs Re: Simplify trig equation No, you want to substitute for $\cot$$\sin^{\small{-1}}(x)$$$ into: $\cot$$2\sin^{\small{-1}}(x)$$=\frac{\cot^2$$\sin^{\small{-1}}(x)$$-1}{2\cot$$\sin^{\small{-1}}(x)$$}$ Using: $\cot$$\sin^{\small{-1}}(x)$$=\frac{\sqrt{1-x^2}}{x}$
October 6th, 2012, 12:13 PM   #5
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Re: Simplify trig equation

Quote:
 Originally Posted by MarkFL No, you want to substitute for $\cot$$\sin^{\small{-1}}(x)$$$ into: $\cot$$2\sin^{\small{-1}}(x)$$=\frac{\cot^2$$\sin^{\small{-1}}(x)$$-1}{2\cot$$\sin^{\small{-1}}(x)$$}$ Using: $\cot$$\sin^{\small{-1}}(x)$$=\frac{\sqrt{1-x^2}}{x}$
Got it thank you very much

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